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In other words is it possible, at least for certain λ and ,
to have matrix multiplication be the same as just multiplying the vector by a
constant? Of course, we probably
wouldn’t be talking about this if the answer was no. So, it is possible for this to happen,
however, it won’t happen for just any value of λ or . If we do happen to have a λ and for which this works (and they will always
come in pairs) then we call λ an eigenvalue
of A and an eigenvector
of A.

So, how do we go about find the eigenvalues and eigenvectors
for a matrix? Well first notice that if then (1) is
going to be true for any value of λ and so we are going to make the assumption
that . With that out of the way let’s rewrite (1)
a little.

Notice that before we factored out the we added in the appropriately sized identity
matrix. This is equivalent to
multiplying things by a one and so doesn’t change the value of anything. We needed to do this because without it we
would have had the difference of a matrix, A,
and a constant, λ, and this can’t be done. We now have the difference of two matrices of
the same size which can be done.

is equivalent to (1). In order to find the eigenvectors for a
matrix we will need to solve a homogeneous system. Recall the fact from the previous section that we
know that we will either have exactly one solution () or we will have infinitely many
nonzero solutions. Since we’ve already
said that don’t want this means that we want the second case.

Knowing this will allow us to find the eigenvalues for a
matrix. Recall from this fact that we
will get the second case only if the matrix in the system is singular. Therefore we will need to determine the
values of λ for which we get,

Once we have the eigenvalues we can then go back and
determine the eigenvectors for each eigenvalue.
Let’s take a look at a couple of quick facts about eigenvalues and eigenvectors.

Fact

If A is an n x n matrix then is an nth
degree polynomial. This polynomial is
called the characteristic polynomial.

To find eigenvalues of a matrix all we need to do is solve a
polynomial. That’s generally not too bad
provided we keep n small. Likewise this fact also tells us that for an n x n
matrix, A, we will have n eigenvalues if we include all repeated
eigenvalues.

Fact

If is the complete list of eigenvalues
for A (including all repeated
eigenvalues) then,

1.If λ occurs only once in the list then we call λsimple.

2.If λ occurs k>1
times in the list then we say that λ has multiplicity
k.

3.If () are
the simple eigenvalues in the list with corresponding eigenvectors ,
,
…, then the eigenvectors are all linearly independent.

4.If λ is an eigenvalue of multiplicity k > 1 then λ will have anywhere from 1 to k linearly independent eigenvectors.

The usefulness of these facts will become apparent when we
get back into differential equations since in that work we will want linearly
independent solutions.

Let’s work a couple of examples now to see how we actually
go about finding eigenvalues and eigenvectors.

Example 1 Find
the eigenvalues and eigenvectors of the following matrix.

Solution

The first thing that we need to do is find the
eigenvalues. That means we need the
following matrix,

In particular we need to determine where the determinant
of this matrix is zero.

So, it looks like we will have two simple eigenvalues for
this matrix, and . We will now need to find the eigenvectors
for each of these. Also note that
according to the fact above, the two eigenvectors should be linearly
independent.

To find the eigenvectors we simply plug in each eigenvalue
into (2)
and solve. So, let’s do that.

:

In this case we need to solve the following system.

Recall that officially to solve this system we use the
following augmented matrix.

Upon reducing down we see that we get a single equation

that will yield an infinite number of solutions. This is expected behavior. Recall that we picked the eigenvalues so
that the matrix would be singular and so we would get infinitely many
solutions.

Notice as well that we could have identified this from the
original system. This won’t always be
the case, but in the 2 x 2 case we can see from the system that one row will
be a multiple of the other and so we will get infinite solutions. From this point on we won’t be actually
solving systems in these cases. We
will just go straight to the equation and we can use either of the two rows
for this equation.

Now, let’s get back to the eigenvector, since that is what
we were after. In general then the
eigenvector will be any vector that satisfies the following,

To get this we used the solution to the equation that we
found above.

We really don’t want a general eigenvector however so we
will pick a value for to get a specific eigenvector. We can choose anything (except ), so pick something that will make
the eigenvector “nice”. Note as well
that since we’ve already assumed that the eigenvector is not zero we must
choose a value that will not give us zero, which is why we want to avoid
in this case. Here’s the eigenvector for this eigenvalue.

Now we get to do this all over again for the second
eigenvalue.

:

We’ll do much less work with this part than we did with
the previous part. We will need to
solve the following system.

Clearly both rows are multiples of each other and so we
will get infinitely many solutions. We
can choose to work with either row.
We’ll run with the first because to avoid having too many minus signs
floating around. Doing this gives us,

Note that we can solve this for either of the two
variables. However, with an eye
towards working with these later on let’s try to avoid as many fractions as
possible. The eigenvector is then,

Summarizing we have,

Note that the two eigenvectors are linearly independent as
predicted.

Example 2 Find
the eigenvalues and eigenvectors of the following matrix.

Solution

This matrix has fractions in it. That’s life so don’t get excited about
it. First we need the eigenvalues.

So, it looks like we’ve got an eigenvalue of multiplicity
2 here. Remember that the power on the
term will be the multiplicity.

Now, let’s find the eigenvector(s). This one is going to be a little different
from the first example. There is only
one eigenvalue so let’s do the work for that one. We will need to solve the following system,

So, the rows are multiples of each other. We’ll work with the first equation in this
example to find the eigenvector.

Recall in the last example we decided that we wanted to
make these as “nice” as possible and so should avoid fractions if we
can. Sometimes, as in this case, we
simply can’t so we’ll have to deal with it.
In this case the eigenvector will be,

Note that by careful choice of the variable in this case
we were able to get rid of the fraction that we had. This is something that in general doesn’t
much matter if we do or not. However,
when we get back to differential equations it will be easier on us if we
don’t have any fractions so we will usually try to eliminate them at this
step.

Also in this case we are only going to get a single
(linearly independent) eigenvector. We
can get other eigenvectors, by choosing different values of . However, each of these will be linearly
dependent with the first eigenvector.
If you’re not convinced of this try it. Pick some values for and get a different vector and check to see
if the two are linearly dependent.

Recall from the fact above that an eigenvalue of
multiplicity k will have anywhere
from 1 to k linearly independent
eigenvectors. In this case we got
one. For most of the 2 x 2 matrices
that we’ll be working with this will be the case, although it doesn’t have to
be. We can, on occasion, get two.

Example 3 Find
the eigenvalues and eigenvectors of the following matrix.

Solution

So, we’ll start with the eigenvalues.

This doesn’t factor, so upon using the quadratic formula
we arrive at,

In this case we get complex eigenvalues which are
definitely a fact of life with eigenvalue/eigenvector problems so get used to
them.

Finding eigenvectors for complex eigenvalues is identical
to the previous two examples, but it will be somewhat messier. So, let’s do that.

:

The system that we need to solve this time is

Now, it’s not super clear that the rows are multiples of
each other, but they are. In this case
we have,

This is not something that you need to worry about, we
just wanted to make the point. For the
work that we’ll be doing later on with differential equations we will just
assume that we’ve done everything correctly and we’ve got two rows that are
multiples of each other. Therefore,
all that we need to do here is pick one of the rows and work with it.

We’ll work with the second row this time.

Now we can solve for either of the two variables. However, again looking forward to
differential equations, we are going to need the “i” in the numerator so solve the equation in such a way as this
will happen. Doing this gives,

So, the eigenvector in this case is

As with the previous example we choose the value of the
variable to clear out the fraction.

Now, the work for the second eigenvector is almost
identical and so we’ll not dwell on that too much.

:

The system that we need to solve here is

Working with the second row again gives,

The eigenvector in this case is

Summarizing,

There is a nice fact that we can use to simplify the work
when we get complex eigenvalues. We need
a bit of terminology first however.

If we start with a complex number,

then the complex conjugate of z is

To compute the complex conjugate of a complex number we
simply change the sign on the term that contains the “i”. The complex conjugate of
a vector is just the conjugate of each of the vector’s components.

We now have the following fact about complex eigenvalues and
eigenvectors.

Fact

If A is an n x n matrix with only real numbers and if is an eigenvalue with eigenvector . Then is also an eigenvalue and its eigenvector is
the conjugate of .

This fact is something that you should feel free to use
as you need to in our work.

Now, we need to work one final eigenvalue/eigenvector
problem. To this point we’ve only worked
with 2 x 2 matrices and we should work at least one that isn’t 2 x 2. Also, we need to work one in which we get an
eigenvalue of multiplicity greater than one that has more than one linearly
independent eigenvector.

Example 4 Find
the eigenvalues and eigenvectors of the following matrix.

Solution

Despite the fact that this is a 3 x 3 matrix, it still
works the same as the 2 x 2 matrices that we’ve been working with. So, start with the eigenvalues

So, we’ve got a simple eigenvalue and an eigenvalue of
multiplicity 2. Note that we used the
same method of computing the determinant of a 3 x 3 matrix that we used in
the previous section. We just didn’t
show the work.

Let’s now get the eigenvectors. We’ll start with the simple eigenvector.

:

Here we’ll need to solve,

This time, unlike the 2 x 2 cases we worked earlier, we
actually need to solve the system. So
let’s do that.

Going back to equations gives,

So, again we get infinitely many solutions as we should
for eigenvectors. The eigenvector is
then,

Now, let’s do the other eigenvalue.

:

Here we’ll need to solve,

Okay, in this case is clear that all three rows are the
same and so there isn’t any reason to actually solve the system since we can
clear out the bottom two rows to all zeroes in one step. The equation that we get then is,

So, in this case we get to pick two of the values for free
and will still get infinitely many solutions.
Here is the general eigenvector for this case,

Notice the restriction this time. Recall that we only require that the
eigenvector not be the zero vector.
This means that we can allow one or the other of the two variables to
be zero, we just can’t allow both of them to be zero at the same time!

What this means for us is that we are going to get two
linearly independent eigenvectors this time.
Here they are.

Now when we talked about linear independent vectors in the
last section we only looked at n vectors each with n components. We can still talk about linear independence
in this case however. Recall back with we did linear independence for functions
we saw at the time that if two functions were linearly dependent then they
were multiples of each other. Well the
same thing holds true for vectors. Two
vectors will be linearly dependent if they are multiples of each other. In this case there is no way to get by multiplying by a constant. Therefore, these two vectors must be
linearly independent.

So, summarizing up, here are the eigenvalues and
eigenvectors for this matrix